Solution:

Solve [latex]2^{3x} = 16^{1-x}[/latex] for [latex]x[/latex].

16 is a power of 2, so we can rewrite [latex]2^{3x} = 16^{1-x}[/latex] as [latex]2^{3x} = \left(2^4\right)^{1-x}[/latex]. Using properties of exponents, we get [latex]2^{3x} = 2^{4(1-x)}[/latex].

Using the one-to-one property of exponential functions, we get
\[ \begin{array}{rcl} 3x & = & 4(1-x) \\
3x & = & 4 – 4x \\
7x & = & 4 \\ & & \\
x & = & \dfrac{4}{7}
\end{array} \]

Graphing [latex]f(x) = 2^{3x}[/latex] (in red) and [latex]g(x) = 16^{1-x}[/latex] (in blue) and see that they intersect at [latex]x =\dfrac{4}{7} \approx 0.571[/latex].

Solution:

Solve [latex]2000 = 1000 \cdot 3^{-0.1 t}[/latex] for [latex]t[/latex].

We begin solving [latex]2000 = 1000 \cdot 3^{-0.1 t}[/latex] by dividing both sides by 1000 to isolate the exponential which yields [latex]3^{-0.1t} = 2[/latex].

As it is inconvenient to write 2 as a power of 3, we use the natural log to get
\[ \begin{array}{rclr} \ln\left(3^{-0.1t}\right) & = & \ln(2) & \\
-0.1 t \ln(3) & = & \ln(2) & \text{Power Rule} \\ & & & \\
t & = & -\dfrac{\ln(2)}{0.1 \ln(3)} & \\ & & & \\
& = & -\dfrac{10\ln(2)}{\ln(3)}
\end{array} \]

We see the graphs of [latex]f(t) = 2000[/latex] (in red) and [latex]g(t) = 1000 \cdot 3^{-0.1 t}[/latex] (in blue) intersect at [latex]t = -\frac{10\ln(2)}{\ln(3)} \approx -6.309[/latex].

Solution:

Solve [latex]9 \cdot 3^{x} = 7^{2x}[/latex] for [latex]x[/latex].

We first note that we can rewrite the equation [latex]9 \cdot 3^{x} = 7^{2x}[/latex] as [latex]3^2 \cdot 3^x = 7^{2x}[/latex] to obtain [latex]3^{x+2} = 7^{2x}[/latex].

As it is not convenient to express both sides as a power of 3 (or 7 for that matter) we use the natural log: [latex]\ln\left(3^{x+2}\right) = \ln\left(7^{2x}\right)[/latex].

The power rule gives [latex](x+2) \ln(3) = 2x \ln(7)[/latex]. Even though this equation appears very complicated, keep in mind that [latex]\ln(3)[/latex] and [latex]\ln(7)[/latex] are just constants.

The equation [latex](x+2) \ln(3) = 2x \ln(7)[/latex] is actually a linear equation (do you see why?) and as such we gather all of the terms with [latex]x[/latex] on one side, and the constants on the other. We then divide both sides by the coefficient of [latex]x[/latex], which we obtain by factoring.
\[ \begin{array}{rclr}
(x+2) \ln(3) & = & 2x \ln(7) & \\
x \ln(3) + 2 \ln(3) & = & 2x \ln(7) & \\
2 \ln(3) & = & 2x \ln(7) – x \ln(3) & \\
2 \ln(3) & = & x (2 \ln(7) – \ln(3)) & \text{Factor}\\ & & & \\
x & = & \dfrac{2 \ln(3)}{2\ln(7) – \ln(3)} & \\
\end{array}\]

We see the graphs of [latex]f(x) = 9 \cdot 3^{x}[/latex] (in red) and [latex]g(x) = 7^{2x}[/latex] (in blue) intersect at [latex]x = \dfrac{2 \ln(3)}{2\ln(7) - \ln(3)} \approx 0.787[/latex].

Solution:

Solve [latex]75 = \dfrac{100}{1 + 3e^{-2t}}[/latex] for [latex]t[/latex].

Our objective in solving [latex]75 = \dfrac{100}{1 + 3e^{-2t}}[/latex] is to first isolate the exponential.

To that end, we clear denominators and get [latex]75\left(1 + 3e^{-2t}\right) = 100[/latex], or
\[ \begin{array}{rclr} 75 + 225e^{-2t} & = & 100 & \\
225e^{-2t} & = & 25 & \\ & & & \\
e^{-2t} & = & \dfrac{1}{9} & \\ & & & \\
\ln\left(e^{-2t}\right) & = & \ln\left( \dfrac{1}{9} \right) & \text{Apply the natural log} \\ & & & \\
-2t & = & -\ln(9) & \text{Power Rule} \\
t & = & \dfrac{\ln(9)}{2} & \\ & & & \\
t & = & \ln(3) & \end{array} \]

To check, we see the graphs of [latex]f(t) = 75[/latex] (in red) and [latex]g(t) = \dfrac{100}{1 + 3e^{-2t}}[/latex] (in blue), intersect at [latex]t =\ln(3) \approx 1.099[/latex].

Solution:

Solve [latex]25^{x} = 5^{x} + 6[/latex] for [latex]x[/latex].

We start solving [latex]25^{x} = 5^{x} + 6[/latex] by rewriting [latex]25 = 5^2[/latex] so that we have [latex]\left(5^2\right)^{x} = 5^{x} + 6[/latex], or [latex]5^{2x} = 5^{x} + 6[/latex].

Even though we have a common base, having two terms on the right hand side of the equation foils our plan of equating exponents or taking logs.

If we stare at this long enough, we notice that we have three terms with the exponent on one term exactly twice that of another. To our surprise and delight, we have a quadratic in disguise.

Letting [latex]u = 5^{x}[/latex], we have [latex]u^2 = \left(5^{x}\right)^2 = 5^{2x}[/latex] so the equation [latex]5^{2x} = 5^{x} + 6[/latex] becomes [latex]u^2 = u + 6[/latex]. Solving this as [latex]u^2 - u - 6=0[/latex] gives [latex]u = -2[/latex] or [latex]u = 3[/latex]. As [latex]u = 5^{x}[/latex], we have \[5^{x} = -2  \text{ or } 5^{x} = 3 \]

[latex]5^{x} = -2[/latex] has no real solution,^[3] so we focus on [latex]5^{x} = 3[/latex]. It isn’t convenient to express 3 as a power of 5, thus we take natural logs and get [latex]\ln\left(5^{x}\right) = \ln(3)[/latex] so that \[x \ln(5) = \ln(3) \text{ or } x = \dfrac{\ln(3)}{\ln(5)}\]

We see the graphs of [latex]f(x) = 25^{x}[/latex] (in red) and [latex]g(x) = 5^{x} + 6[/latex] (in blue) intersect at [latex]x = \dfrac{\ln(3)}{\ln(5)}[/latex].

Solution:

Solve [latex]\dfrac{e^{x} - e^{-x}}{2} = 5[/latex] for [latex]x[/latex].

Clearing the denominator in [latex]\dfrac{e^{x} - e^{-x}}{2} = 5[/latex] gives [latex]e^{x} - e^{-x} = 10[/latex], at which point we pause to consider how to proceed. Rewriting [latex]e^{-x} = \dfrac{1}{e^{x}}[/latex], we see we have another denominator to clear: [latex]e^{x} - \dfrac{1}{e^{x}} = 10[/latex].

Doing so gives [latex]e^{2x} - 1 = 10e^{x}[/latex], which, once again fits the criteria of being a quadratic in disguise.

If we let [latex]u = e^{x}[/latex], then [latex]u^2 = e^{2x}[/latex] so the equation [latex]e^{2x} - 1 = 10e^{x}[/latex] can be viewed as [latex]u^2-1 = 10u[/latex]. Solving [latex]u^2 - 10u - 1 = 0[/latex] using the quadratic formula gives [latex]u = 5 \pm \sqrt{26}[/latex].

From this, we have [latex]e^{x} = 5 \pm \sqrt{26}[/latex]. Because [latex]5 - \sqrt{26}[/latex] < 0, we get no real solution to [latex]e^{x} = 5 - \sqrt{26}[/latex] (why not?) but for [latex]e^{x} = 5 + \sqrt{26}[/latex], we take natural logs to obtain [latex]x = \ln\left(5 + \sqrt{26}\right)[/latex].

We see the graphs of [latex]f(x) = \dfrac{e^{x} - e^{-x}}{2}[/latex] (in red) and [latex]g(x) = 5[/latex] (in blue) intersect at [latex]x = \ln\left(5 + \sqrt{26}\right) \approx 2.312[/latex].

Solution:

Write a formula for [latex]f^{-1}(x)[/latex].

We start by writing [latex]y=f(x)[/latex], and interchange the roles of [latex]x[/latex] and [latex]y[/latex]. To solve for [latex]y[/latex], we first clear denominators and then isolate the exponential function.
\[ \begin{array}{rclr}
y & = & \dfrac{5e^{x}}{e^{x}+1} & \\ & & & \\
x & = & \dfrac{5e^{y}}{e^{y}+1} & \text{Switch } x \text{ and } y \\ & & & \\
x \left(e^{y}+1\right) & = & 5e^{y} & \\ & & & \\
x e^{y}+x & = & 5e^{y} & \\ & & & \\
x & = & 5e^{y} – x e^{y} & \\ & & & \\
x & = & e^{y}(5 – x) & \\ & & & \\
e^{y}& = & \dfrac{x}{5-x} & \\ & & & \\
\ln\left(e^{y}\right) & = & \ln\left(\dfrac{x}{5-x}\right) & \\ & & & \\
y & = & \ln\left(\dfrac{x}{5-x}\right) & \\
\end{array}\]
We claim [latex]f^{-1}(x) = \ln\left(\dfrac{x}{5-x}\right)[/latex]. To verify this analytically, we would need to verify the compositions [latex]\left(f^{-1} \circ f\right)(x) = x[/latex] for all [latex]x[/latex] in the domain of [latex]f[/latex] and that [latex]\left(f \circ f^{-1}\right)(x) = x[/latex] for all [latex]x[/latex] in the domain of [latex]f^{-1}[/latex]. We leave this, as well as a graphical check, to the reader in Exercise 41.

Solution:

Solve [latex]\dfrac{5e^{x}}{e^{x}+1} = 4[/latex].

We recognize the equation [latex]\dfrac{5e^{x}}{e^{x}+1} = 4[/latex] as [latex]f(x) = 4[/latex]. Hence, our solution is \[ \begin{array}{rcl} x &=& f^{-1}(4) \\ & & \\ &=& \ln\left(\dfrac{4}{5-4}\right) \\& & \\ &=& \ln(4) \end{array} \]

We can check this fairly quickly algebraically. Using [latex]e^{\ln(4)} = 4[/latex], we find [latex]\dfrac{5e^{\ln(4)}}{e^{\ln(4)}+1} = \dfrac{5(4)}{4+1} = \dfrac{20}{5} = 4[/latex].